OK. I tried sth different to rebuild my proof, with all the necessary details and explanation (I believe) contained. Here is the sketch and the complete version:
Sketch of the proof: first show that $f$ can be continuously extended to $0$ on the arc of the sector. Define a new function depending on $f$, possibly by rotating, such that it equals to $0$ on the boundary. Then apply the maximum modulus principle.
Proof: WLOG we assume that $\theta = 0$. $S:= \left\{ z\in\mathbb{D}: 0< \text{arg} z <\varphi\right\}$ is the sector. $f \rightrightarrows 0$ on $S$, i.e. $\forall \epsilon >0$, $\exists \delta >0$, s.t. $\forall z\in S$, and $\forall w\in \partial \mathbb{D}$, whenever $|z-w|<\delta$, we have $\left |f(z) \right |<\epsilon$. Here we explains the uniformity. Basically we can imagine it dies to 0 very well on the boundary.
Now we may extend $f$ to $\partial \mathbb{D}_{(0, \varphi)}$ (the arc of the sector) by letting $f = 0$.
For $\varphi/2$, there is a nature number $N\in \mathbb{N}$ s.t. $N \varphi /2 > 2\pi$. We want to somehow cover the whole boundary of the disc so we may apply the max. mod. principle. However, the interval of argument is open. It is necessary and convenient to choose a number smaller than$\varphi$so that we can rotate it many times to cover all of the boundary with no holes at all. Consider
$$F(z):= f(z)f(e^{-i\varphi/2}z)f(e^{-i2\varphi/2}z)\cdots f(e^{-iN\varphi/2}z)$$
on $\mathbb{D}$. First, it is bounded on $\mathbb{D}$. Next, we show that it can be continuously extended to $\bar{\mathbb{D}}$: $\forall w\in \partial \mathbb{D}$, $\exists k\in \left\{ 0, 1, 2, \cdots, N\right\}$ s.t. $k\varphi/2 < \text{arg} w \leq (k+1)\varphi/2$. Now given $\epsilon >0$, $\exists \delta >0$ s.t. $|z-e^{-ik\varphi/2}w|<\delta$ (Rotate it back to where we start) implies $\left |f(z) \right |<\epsilon$; for the other numbers $j \neq k$ in $\left\{ 0, 1, 2, \cdots, N\right\}$, $\left |f(e^{-ij\varphi/2}z) \right |<B$ as $f$ is bounded. Hence
$$\left |F(z) \right |= \prod_{j=0}^{N}\left |f(e^{-ij\varphi/2}z) \right |<B^N\epsilon.$$
So we may define $F$ to be $0$ on $\partial \mathbb{D}$. Apply the maximum modulus principle. We conclude $F \equiv 0$ on $\bar{\mathbb{D}}$.
If $f \not \equiv 0$, then $f$ has at most countably many roots since it is analytic. We use this fact to conclude a contradiction. Now fix an arbitrary $\alpha$, $\forall re^{i\alpha}\in \mathbb{D}$, at least one of the term in $F(\alpha) = 0$ is zero. This gives a root for each $r \in (0, 1)$. But $(0, 1)$ is uncountable. Contradiction.
Therefore $f\equiv 0$. $\square $
Best Answer
Here's one way to do it. Let $M$ be a bound for $|f|$ on the unit disc and let $t$ be slightly smaller than $\varphi-\theta$ and define $$ g(z) = f(z)f(ze^{it})f(ze^{2it})\cdots f(ze^{nit}) $$ where $n$ is chosen so large that $nt > 2\pi$. Let $\varepsilon > 0$. By assumption there is an $r < 1$ such that $|f(z)| < \varepsilon$ for $r < |z| < 1$ and $\theta < \arg z < \phi$. Hence $$ |g(z)| < M^n \varepsilon $$ for all $z$ with $r < |z| < 1$. (One factor has modulus less than $\varepsilon$ and the other factors less than $M$.) By the maximum modulus principle, $|g| < M^n\varepsilon$ on the whole disc, and since $\varepsilon$ was arbitrary, we must have that $g(z) = 0$ for all $z$ in the unit disc.
Hence one of the factors, and consequently all factors, of $g$ vanishes identically (otherwise $g$ would have at most countably many zeros).